Consider the following :
1. `x+x^(2)` is continuous at x=0
2. `x+cos(1/x)` is discontinuous at x=0
3. `x^(2)+cos(1/x)` is continuous at x=0`
Which of the above are correct?

Options

A. 1 and 2 olny
B. 2 and 3 only
C. 1 and 3 only
D. 1, 2 and 3

Explanation

To determine which of the statements about continuity at \( x = 0 \) are correct, we will analyze each function step by step.

### Step 1: Analyze \( f(x) = x + x^2 \)

1. **Function Definition**: The function \( f(x) = x + x^2 \) is a polynomial function.
2. **Continuity of Polynomial Functions**: Polynomial functions are continuous everywhere on the real number line.
3. **Evaluate at \( x = 0 \)**: Since \( f(x) \) is continuous for all \( x \), it is also continuous at \( x = 0 \).

**Conclusion for Statement 1**: The statement "x + x^2 is continuous at x = 0" is **True**.

### Step 2: Analyze \( g(x) = x + \cos(1/x) \)

1. **Function Definition**: The function \( g(x) = x + \cos(1/x) \) involves a polynomial part \( x \) and a non-polynomial part \( \cos(1/x) \).
2. **Limit as \( x \to 0 \)**:
- **Left-hand limit**: As \( x \) approaches \( 0 \) from the left, \( \cos(1/x) \) oscillates between -1 and 1. Thus, \( \lim_{x \to 0^-} g(x) \) does not exist because it fluctuates.
- **Right-hand limit**: As \( x \) approaches \( 0 \) from the right, \( \cos(1/x) \) also oscillates between -1 and 1. Thus, \( \lim_{x \to 0^+} g(x) \) does not exist for the same reason.
3. **Conclusion**: Since both the left-hand and right-hand limits do not exist, \( g(x) \) is discontinuous at \( x = 0 \).

**Conclusion for Statement 2**: The statement "x + cos(1/x) is discontinuous at x = 0" is **True**.

### Step 3: Analyze \( h(x) = x^2 + \cos(1/x) \)

1. **Function Definition**: The function \( h(x) = x^2 + \cos(1/x) \) includes a polynomial part \( x^2 \) and a non-polynomial part \( \cos(1/x) \).
2. **Limit as \( x \to 0 \)**:
- **Left-hand limit**: As \( x \) approaches \( 0 \) from the left, \( x^2 \) approaches \( 0 \) and \( \cos(1/x) \) oscillates between -1 and 1. Thus, \( \lim_{x \to 0^-} h(x) = 0 + \text{oscillating value} \) which does not converge to a single value.
- **Right-hand limit**: Similarly, as \( x \) approaches \( 0 \) from the right, \( \lim_{x \to 0^+} h(x) = 0 + \text{oscillating value} \) which also does not converge.
3. **Conclusion**: The polynomial part \( x^2 \) approaches \( 0 \), but the oscillation of \( \cos(1/x) \) means that \( h(x) \) does not have a definite limit as \( x \) approaches \( 0 \).

**Conclusion for Statement 3**: The statement "x^2 + cos(1/x) is continuous at x = 0" is **False**.

### Final Conclusion

- Statement 1: True
- Statement 2: True
- Statement 3: False

Thus, the correct statements are 1 and 2.